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#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 43 Maths Textbook Solution.

$\pi$

Hint: Use indefinite formula and put the limits to solve this integral

Given:

$\int_{0}^{4}\frac{1}{\sqrt{4x-x^{2}}}dx$

Solution:

$\int_{0}^{4}\frac{1}{\sqrt{4x-x^{2}}}dx=\int_{0}^{4}\frac{1}{\sqrt-{\left ( x^{2}-2.2x+\left ( 2 \right )^{2}-\left ( 2 \right )^{2} \right )}}dx$

$=\int_{0}^{4}\frac{1}{\sqrt-{\left \{ \left ( x-2 \right )^{2}-4 \right \}}}dx$

$\left [ \because \left ( a-b \right )^{2}=a^{2}-2ab+b^{2} \right ]$

$=\int_{0}^{4} \frac{1}{\sqrt{4-(x-2)^{2}}} d x$

$=\int_{0}^{4} \frac{1}{\sqrt{2^{2}-(x-2)^{2}}} d x\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)\right]$

$=\left[\sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{4}$

$=\left[\sin ^{-1}\left(\frac{4-2}{2}\right)-\sin ^{-1}\left(\frac{0-2}{2}\right)\right]$

$=\left[\sin ^{-1}\left(\frac{2}{2}\right)-\sin ^{-1}\left(\frac{-2}{2}\right)\right]$

$=\left[\sin ^{-1}(1)-\sin ^{-1}(-1)\right]$

$=\left[\sin ^{-1}(1)+\sin ^{-1}(1)\right]\left[\because \sin ^{-1}(-\theta)=-\sin \theta\right]$

$=2 \sin ^{-1}(1)$

$=2.\frac{\pi }{2}$

$=\pi$